# cyclotomic extension, quadratic extension of $K=\mathbb{Q}(\sqrt{17})$ such that $K/\mathbb{Q}$ is cyclic

There exist a quadratic extension $J$ of $K=\mathbb{Q}(i)$ such that the extension $J / \mathbb{Q}$ is cyclic? The same question with $K=\mathbb{Q}(\sqrt{17})$

I did this problem , but only with "luck" because , I know that cyclotomic extensions are cyclic, and the degree of the cyclotomic extension $[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\phi(n)$. So I searched first for some $n$ such that $\phi(n)=4$ , for example $n=8$. But I also need that this extension also contains $i$. I don't know how to check this in general. In this case I realized that $\zeta_8^2 = i$ and I'm done , but for example I don't how to check this in general. For example if $n=5$

And I don't have any idea how to attack the problem with $K=\mathbb{Q}(\sqrt{17})$ This is a problem of cyclotomic extensions, I don't know how to relate cyclotomic extensions with $\sqrt{17}$

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It's known that if $p$ is prime then ${\bf Q}(\zeta_p)$ contains $\sqrt p$ if $p\equiv1\pmod4$, $\sqrt{-p}$ if $p\equiv3\pmod4$. More generally, Kronecker's Jugendtraum, which see, was that every abelian extension is contained in a cyclotomic extension. –  Gerry Myerson Nov 4 '12 at 2:41
You know how can i prove that proposition? –  Daniel Nov 4 '12 at 2:50
Which one? The Jugendtraum is hard, the one about square roots comes from evaluating the Gauss sum $\sum e^{2\pi ix^2/p}$. –  Gerry Myerson Nov 4 '12 at 3:02
Daniel, you should take a look at the following answer by Bill Dubuque to a related question. –  Adrián Barquero Nov 4 '12 at 3:10
I can conclude, using all the results that does not exist a quadratic extension of $K=\mathbb{Q}(\sqrt{17})$ that is also cyclotomic. But maybe there exist some non cyclotomic quadratic extension $J$ of $K$ , such that $J/\mathbb{Q}$ is cyclic. But maybe is a subfield of $\mathbb{Q} (\zeta_{17})$ –  Daniel Nov 4 '12 at 3:30

There is no need to go to algebraic number theory for this. If you know that the discriminant of the polynomial $x^p - 1$ is $(-1)^{p(p-1)/2}p^p$ then this is not so hard. For then $\Bbb{Q}$ adjoined the square root of the discriminant is a degree $2$ extension of the rationals that is contained in $\Bbb{Q}(\zeta_p)$, namely $\Bbb{Q}\left(\sqrt{(-1)^{p(p-1)/2}p}\right)$. If $p \equiv 1 \pmod 4$ then your extension is $\Bbb{Q}(\sqrt{p})$, if $p \equiv 3 \mod 4$ it is $\Bbb{Q}(\sqrt{-p})$. Now $17 \equiv 1 \pmod{4}$ and so you just need to consider a degree 2 extension $J/\Bbb{Q}(\sqrt{17})$ inside of $\Bbb{Q}(\zeta_{17})$. This is because for $p$ prime, $\textrm{Gal}(\Bbb{Q}(\zeta_p)/\Bbb{Q})$ is a cyclic group of order $p-1$.

Added for OP: Your Galois group is cyclic of order 16. For each positive integer dividing 16 there is exactly one subgroup of that order. Now the subgroup $H$ of order (and hence index 4) is generated by $\langle x^4 \rangle$. The subgroup of index $2$ $K$ say is generated by $\langle x^2 \rangle$. Since $H \subseteq K$ by the fundamental theorem of Galois Theory

$$E^K \subseteq E^H,$$

i.e. your extension of degree 4 will containt the unique extension of degree 2, namely $\Bbb{Q}(\zeta_{17})$.

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Thanks for your answer without using too much propositions :D! But it's not finished, because $J=\mathbb{Q}(\zeta_{17} )$ it's not the field that I want, because $[J:\mathbb{Q}] = 16$ and not 4 as I want. But obviously I have to consider a subfield $L$ of $J$ such that $[L:\mathbb{Q}(\sqrt{17})]=2$ And now I have to study a little what are the subfield of cyclotomic extensions –  Daniel Nov 4 '12 at 4:03
@Daniel If you know the fundamental theorem of Galois theory this shouldn't be too hard. I suggest looking at subextensions of the form $\Bbb{Q}(\zeta_{17} + \zeta_{17}^{-1})$ say. –  fpqc Nov 4 '12 at 4:09
A little more complicated than that. I’m preparing a messy answer... –  Lubin Nov 4 '12 at 4:25
@BenjaLim I think that I have to consider the fixed field of the group of four elements –  Daniel Nov 4 '12 at 4:32
Since ${\bf Q}(\zeta_{17})$ is cyclic, it has only one subfield of degree 4, which contains the only subfield of degree 2. –  Gerry Myerson Nov 4 '12 at 5:31
To make things easy for myself, I'm going to call a primitive $17$-th root of unity $z$ instead of $\zeta$. Then $K=\mathbb{Q}(z)$ is of degree $16$ over $\mathbb Q$, and its Galois group is naturally isomorphic to $(\mathbb Z/17\mathbb Z)^*$, which is cyclic, and we can choose $6$ for a generator. We have $6^2\equiv 2\pmod{17}$ and $6^4\equiv 4\pmod{17}$. Applying this to our field extension, the corresponding generator of the Galois group sends $z$ to $z^6$. The (unique) subgroup of order $8$ is generated by $z\mapsto z^2$, and the subgroup of order $4$ is generated by $z\mapsto z^4$. These groups are of index $2$ and $4$, respectively, and their fixed fields are the (unique) quadratic and quartic extensions of $\mathbb Q$ contained in $K$.
Now comes the messy computational part. You hope (and it usually seems to work) that the trace of $z$ down to each subfield will actually be a generator of the field in question. So you hope that $r=z+z^2+z^4+z^8+z^{-1}+z^{-2}+z^{-4}+z^{-8}$ will generate the quadratic extension and $x=z+z^4+z^{-1}+z^{-4}$ will generate the quartic extension. I confess that I used machine computation, but it shouldn’t be beyond handwork to see what I found: $r^2+r-4=0$. So $r=(-1\pm\sqrt{17})/2$. Yay! Next I calculated $x^2$ and $rx$, and it’s a stroke of luck that $x^2=rx+1$ !! So $x=(r\pm\sqrt{8-r})/2$.
If we take $r=(-1-\sqrt{17})/2$, then $8-r=(17+\sqrt{17})/2$, so the above computation shows that the quartic cyclic extension is $\mathbb Q\left(\sqrt{\frac{17+\sqrt{17}}2}\;\right)$, but I have to confess that I’ve run out of steam in the business of describing the action of the Galois group.